[Swift]LeetCode1172. 餐盤棧 | Dinner Plate Stacks

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You have an infinite number of stacks arranged in a row and numbered (left to right) from 0, each of the stacks has the same maximum capacity.

Implement the DinnerPlates class:

• DinnerPlates(int capacity) Initializes the object with the maximum capacity of the stacks.
• void push(int val) pushes the given positive integer val into the leftmost stack with size less than capacity.
• int pop() returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns -1 if all stacks are empty.
• int popAtStack(int index) returns the value at the top of the stack with the given index and removes it from that stack, and returns -1 if the stack with that given index is empty.

Example:

Input: 
["DinnerPlates","push","push","push","push","push","popAtStack","push","push","popAtStack","popAtStack","pop","pop","pop","pop","pop"]
[[2],[1],[2],[3],[4],[5],[0],[20],[21],[0],[2],[],[],[],[],[]]
Output: 
[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]

Explanation: 
DinnerPlates D = DinnerPlates(2);  // Initialize with capacity = 2
D.push(1);
D.push(2);
D.push(3);
D.push(4);
D.push(5);         // The stacks are now:  2  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 2.  The stacks are now:     4
                                                       1  3  5
                                                       ﹈ ﹈ ﹈
D.push(20);        // The stacks are now: 20  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.push(21);        // The stacks are now: 20  4 21
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 20.  The stacks are now:     4 21
                                                        1  3  5
                                                        ﹈ ﹈ ﹈
D.popAtStack(2);   // Returns 21.  The stacks are now:     4
                                                        1  3  5
                                                        ﹈ ﹈ ﹈ 
D.pop()            // Returns 5.  The stacks are now:      4
                                                        1  3 
                                                        ﹈ ﹈  
D.pop()            // Returns 4.  The stacks are now:   1  3 
                                                        ﹈ ﹈   
D.pop()            // Returns 3.  The stacks are now:   1 
                                                        ﹈   
D.pop()            // Returns 1.  There are no stacks.
D.pop()            // Returns -1.  There are still no stacks.

Constraints:

• 1 <= capacity <= 20000
• 1 <= val <= 20000
• 0 <= index <= 100000
• At most 200000 calls will be made to push, pop, and popAtStack.

---

咱們把無限數量 ∞ 的棧排成一行，按從左到右的次序從 0 開始編號。每一個棧的的最大容量 capacity 都相同。

實現一個叫「餐盤」的類 DinnerPlates：

• DinnerPlates(int capacity) - 給出棧的最大容量 capacity。
• void push(int val) - 將給出的正整數 val 推入 從左往右第一個 沒有滿的棧。
• int pop() - 返回 從右往左第一個 非空棧頂部的值，並將其從棧中刪除；若是全部的棧都是空的，請返回 -1。
• int popAtStack(int index) - 返回編號 index 的棧頂部的值，並將其從棧中刪除；若是編號 index 的棧是空的，請返回 -1。

示例：

輸入： 
["DinnerPlates","push","push","push","push","push","popAtStack","push","push","popAtStack","popAtStack","pop","pop","pop","pop","pop"]
[[2],[1],[2],[3],[4],[5],[0],[20],[21],[0],[2],[],[],[],[],[]]
輸出：
[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]

解釋：
DinnerPlates D = DinnerPlates(2);  // 初始化，棧最大容量 capacity = 2
D.push(1);
D.push(2);
D.push(3);
D.push(4);
D.push(5);         // 棧的現狀爲：    2  4
                                    1  3  5
                                    ﹈ ﹈ ﹈
D.popAtStack(0);   // 返回 2。棧的現狀爲：      4
                                          1  3  5
                                          ﹈ ﹈ ﹈
D.push(20);        // 棧的現狀爲：  20  4
                                   1  3  5
                                   ﹈ ﹈ ﹈
D.push(21);        // 棧的現狀爲：  20  4 21
                                   1  3  5
                                   ﹈ ﹈ ﹈
D.popAtStack(0);   // 返回 20。棧的現狀爲：       4 21
                                            1  3  5
                                            ﹈ ﹈ ﹈
D.popAtStack(2);   // 返回 21。棧的現狀爲：       4
                                            1  3  5
                                            ﹈ ﹈ ﹈ 
D.pop()            // 返回 5。棧的現狀爲：        4
                                            1  3 
                                            ﹈ ﹈  
D.pop()            // 返回 4。棧的現狀爲：    1  3 
                                           ﹈ ﹈   
D.pop()            // 返回 3。棧的現狀爲：    1 
                                           ﹈   
D.pop()            // 返回 1。如今沒有棧。
D.pop()            // 返回 -1。仍然沒有棧。

提示：

• 1 <= capacity <= 20000
• 1 <= val <= 20000
• 0 <= index <= 100000
• 最多會對 push，pop，和 popAtStack 進行 200000 次調用。

---

Runtime: 2024 ms

Memory Usage: 45 MB

 1 class DinnerPlates {
 2     var map:[Int:[Int]] = [Int:[Int]]()
 3     var cap:Int = 0
 4     var curr:Int = 0
 5     var last:Int = 0
 6     var count:Int = 0
 7 
 8     init(_ capacity: Int) {
 9         self.cap = capacity
10         map[curr] = [Int]()        
11     }
12     
13     func push(_ val: Int) {
14         while(map[curr] != nil  && map[curr,default:[Int]()].count == cap)
15         {
16             curr += 1
17         }
18         map[curr,default:[Int]()].append(val)
19         last = max(last, curr)
20         count += 1
21         
22     }
23     
24     func pop() -> Int {
25         if count == 0 {return -1}
26         while(last>=0 && map[last,default:[Int]()].isEmpty)
27         {
28             last -= 1
29         }
30         count -= 1
31         curr = min(curr, last)
32         return map[last,default:[Int]()].removeLast()
33     }
34     
35     func popAtStack(_ index: Int) -> Int {
36         if(map[index] == nil || map[index,default:[Int]()].isEmpty)
37         {
38             return -1
39         }
40         count -= 1
41         curr = min(curr, index)
42         return map[index,default:[Int]()].removeLast()
43     }
44 }
45 
46 /**
47  * Your DinnerPlates object will be instantiated and called as such:
48  * let obj = DinnerPlates(capacity)
49  * obj.push(val)
50  * let ret_2: Int = obj.pop()
51  * let ret_3: Int = obj.popAtStack(index)
52  */